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Discretization replaces a continuous mathematical problem by equations for finitely many unknowns. A spatial grid samples a function; a finite element method instead chooses coefficients in a finite-dimensional trial space. A time discretization supplies update equations at selected times. These choices specify what is computed, before any roundoff error is introduced.

For the heat equation $u_t=u_{xx}$, a uniform spatial grid $x_j=jh$ givesThis is a semidiscretization: time remains continuous. Applying the Forward Euler method gives a fully discrete updateFor a smooth exact solution, the normalized local truncation error is $O(k+h^2)$. Convergence additionally requires stability of a numerical method; for this one-dimensional update, $0\le k/h^2\le 1/2$ makes each new value a convex combination of neighboring values. Spatial and temporal refinement therefore interact, rather than giving two independent accuracy knobs.

For a smooth solution of $y^{\prime }=f(x,y)$, Taylor's theorem gives the displayed local truncation error when an Euler method step starts at an exact solution value. Thus the one-step defect is $O(h^2)$. Dividing it by $h$ gives a normalized local truncation error of $O(h)$; these are different conventions for reporting the same residual. Also $y^{\prime \prime }=f_x+f_{y}f$ along the solution by the chain rule. A global $O(h)$ error conclusion additionally requires the usual regularity and stability control over successive steps.

For an exact solution $y^{\prime }=f(t,y)$, define the normalized local truncation error of the implicit midpoint rule byIf $\Vert y^{\prime \prime }\Vert \le M_2$, $\Vert y^{\prime \prime \prime }\Vert \le M_3$ and the state Jacobian matrix of $f$ has operator norm at most $L$ on the intervening state segments, then the displayed bound holds uniformly on a fixed finite interval. To prove it, put $t=t_n+h/2$. Symmetric Taylor expansions giveFor the first bound, integrate the second-order remainder of $y^{\prime }(t+s)$ over the symmetric step; for the second, average the second-order remainders of $y$ at the endpoints. The odd terms cancel. The Lipschitz condition for $f$ and $y^{\prime }(t)=f(t,y(t))$ finish the proof by the triangle inequality. The unnormalized step residual is $h{\tau }_n=O(h^3)$.

There is a genuine indexing defect in the printed PDF: its new value is at $m+1$, while every old-time coefficient is centered at $m$. On a fixed grid the literal residual containsFor a general solution the leading term is $u_x/(\mu \delta )$, so this is not a consistent diffusion discretization. It also leaves one new interior value unspecified while imposing a generally incompatible update on a prescribed boundary value. The calculation that follows is for the intended same-node update $U_m^{n+1}=U_m^n+k{L}_{\delta }{U}_m^n$; that correction is necessary, not an unnoticed transcription change.

For the midpoint flux stencil for one-dimensional diffusion, defineAssume sufficient smoothness of $a$ and the solution for the displayed remainders. Multiplying the centered Taylor expansions of $a$ and $u$ givesThe first time difference is $(u(t+k)-u(t))/k=u_t+k{u}_{tt}/2+O(k^2)$. After using the diffusion equation, the normalized local truncation error isTherefore, for the corrected update,The first is the residual divided by the time step; the second is the unnormalized one-step defect. At fixed $\mu =k/{\delta }^2$ they are respectively $O({\delta }^2)$ and $O({\delta }^4)$. A rate based on derivatives is not justified for a merely bounded nonsmooth $a$.

Write $h=\Delta x$, $k=\Delta t$, and assume $a,u$ are sufficiently smooth for the following Taylor expansions. The symmetric half-grid diffusion consistency calculation giveswhereThe normalized local truncation error of the full update is consequentlyThe PDF prints an advection equation, $u_t=a(x)u_x$, whereas this stencil approximates conservative diffusion, $u_t=(a{u}_x{)}_x$. For the literal printed equation the leading defect is $(a-a^{\prime })u_x-a{u}_{xx}$, which generally does not vanish: the method is inconsistent, with an $O(1)$ normalized defect.

For the intended variable-coefficient conservative diffusion equation, the first two terms cancel and ${\mathcal{T}}_{h,k}=O(k+h^2)$. Under the parabolic step scaling $k=O(h^2)$, this is $O(h^2)$, while the unnormalized one-step local truncation error is $k{\mathcal{T}}_{h,k}=O(h^4)$. These two orders use different normalizations. With numerical stability and compatible smooth initial and boundary conditions, the intended diffusion problem has global error $O(k+h^2)$ on a fixed time interval. Positivity and boundedness of $a$ alone do not justify the displayed Taylor expansion; coefficient regularity is also needed for that error formula.

Here $y^{\prime }=x/(1+y^2)$ and $y(0)=1$. Since $x\ge 0$, the solution is nondecreasing, so $y(x)\ge 1$. In fact $0\le y^{\prime }\le 1/2$ on $[0,1]$, which also bounds the solution and ensures it extends across this entire interval. The smooth right-hand side makes $y$ twice continuously differentiable, with second derivativeFor $y\ge 1$ and $0\le x\le 1$,so $y^{\prime \prime }\ge 0$. The subtracted term in the first expression is nonnegative, giving the useful boundThe integral form of the Taylor remainder now givesTherefore the normalized local truncation error satisfies the stronger estimateThis proves the requested bound for $n=1,\dots ,N-1$, and also for $n=0$, which will be needed to start the global error estimate. For $N=1$ the stated range is empty, but the estimate at $n=0$ still applies.

Work on a fixed finite interval $[a,a+T]$. Write $Y_n=y(x_n)$, $t=x_n+h/2$, and use the paper's normalized local truncation errorAssume $y$ has bounded continuous derivatives through order three on this interval, with $\Vert y^{\prime \prime }\Vert \le M_2$ and $\Vert y^{\prime \prime \prime }\Vert \le M_3$. Assume also that the Jacobian matrix $f_y$ has operator norm at most $L$ on the segments joining $y(t)$ to the endpoint averages. For a scalar equation this last hypothesis is simply $|\partial f/\partial y|\le L$ there.

Two symmetric estimates separate the errors. By Taylor's theorem applied to $y^{\prime }$ about $t$, the linear term integrates to zero on $[-h/2,h/2]$, givingLikewise the two linear terms in the endpoint expansions cancel, and their second-order remainders giveThe derivative bound on $f$ gives a Lipschitz condition in its second variable. Since $y^{\prime }(t)=f(t,y(t))$, the triangle inequality now yieldsThe constant is independent of $n$ and $h$ as long as the steps remain in the fixed interval. This implicit midpoint consistency bound proves numerical consistency of order two. The corresponding unnormalized residual is $h{\tau }_n=O(h^3)$; the two powers conventionally called local truncation error differ because the paper divides by the step size. If local error instead means the error of an exactly solved numerical step started at $Y_n$, the small-step Lipschitz condition gives the bound $h\Vert {\tau }_n\Vert /(1-hL/2)=O(h^3)$ as well.

Define the one-step local truncation error as the defect obtained by inserting the exact solution into an update:For a smooth solution, Taylor's theorem and $y^{\prime }=f(x,y)$ giveThus $m=2$ for the one-step defect convention. The leading derivative can also be written $y^{\prime \prime }=f_x+f_{y}f$ by the chain rule.

Some conventions instead call ${\tau }_{n+1}={\delta }_{n+1}/h$ the truncation error. That normalized local truncation error is $O(h)$, so $m=1$ under the normalized convention. Both describe the same Euler one-step defect; neither should be confused with an unqualified global error claim.